Question

A 3-phase 60 Hz 50 km transmission line delivers 20 MW of power to a load at 69 kV and a power factor of 0.8 lagging. The line has the following parameters r = 0.1112/km L = 1.11 mH/km C = negligible Determine: The line impedance. (4 Marks) The "receiving end" phase voltage and current (7 Marks) The "sending end" voltage and current (10 Marks) The voltage regulation. (4 Marks)

Answer 1

lagging * Given that a 3-0, 60HZ, 60km line P = 20m R = 69 kV cosp= 0.8 Se olle/km L=1.11 mH 1 km (is we have to find line impedence line Impedence is Given by z= Reix R = 0.11 x60 (for 60 km] R= 6.602 X = WL*60 = 217FxLX50 = 2x3-14 x 60 x 1.1 x60x16 x = 25.094 R line Impedence = 6.6+ / 25.09 r. We have find the Receiving end phase voltage & Current. As Giren VR = 69 kV line for Uprave v - 39.83 kw Phase (R = 39.83 kv | Receiving Current P = n3 v L coso 20x106 = 23 x 69 x103 x 4, x 0.8 f = 209.18 A

We have to find sending and voltage & current In the short transmission line sending and and receiving end current are same so Is = 209.18A & only the Magnituse & for sending and voltage V= R + FRZ (Phai) (phase) A 64 X10% + 209-18 L-36.96 (66+25-07 Vs = 69 x10% + 209 L-36-86. [6.6+ 25-091 ] V = 44.214 L4.36 KV . (Phase) Phase) (line) Vs = 3x Vsrom. = 3x44.214 _4-36 s = 76.35 [4.36 kv. We have to find the voltage regulation Voltage Regulation = V-VR x 100 * is given by 76-35- 69 x100 76.35 TV.R = 9.62 %